Contractions - find $f$ such that $f:(0,1) \to (0,1)$ with no fixed points The given solution I have is that 
$f(x) = 0.5 + 0.5x$ but I am not entirely how these results are derived.
A function $f : X \to X $ is a contraction if $d(f(x),f(y)) \leq \gamma d(x,y)$
so here,
$$|f(x) - f(y)| \leq \gamma (x-y)$$
$$|0.5 + 0.5x - 0.5 -0.5y| \leq 0.5(x-y)$$
So obviously the solution checks out since $\gamma = 0.5 < 1$
We could have also chosen $f(x) = c + cx$ where $0<c<1$ , correct?
How would my approach have changed if we had to find
$a)$ A contraction $f:(0,1) \to (0,1)$ with at least 1 fixed point
$b)$ A contraction $f:(0,5) \to (0,5)$ with no fixed points
 A: First, is there any result that guarantees the existence of an application without fixed points?

 Yes, fixed-point theorems requires a compact set, while (0,1) is bounded but not closed. Thus, there exists a continuous application without fixed points.

Then, you must adhere to some requirements:


*

*Your application should be a contraction, the definition is the one you said;

*It must return values in $(0,1$), with no fixed points.


Having not to deal with $0$ and $1$ (the toughest ones), you can say that $f(x) = \gamma x$, such that $0 < \gamma < 1$ satisfies the requests. Also, shifting by a constant $\phi$ such that $\phi + \gamma \le 1$ gives you acceptable results. So, every function written as $f(x) = \phi + \gamma x$ is a contraction with no fixed points in $(0,1)$.
How about a contraction with at least one fixed point?

 Identity doesn't work well here, since $\gamma = 1$ doesn't give a contraction. What if you try to contract the unit open interval to a point which belongs to $(0,1)$?

What changes if we search a contraction $(0,5) \to (0,5)$ without fixed points?

 You can partly recycle the previous reasonings. However, while $0 < \gamma < 1$ still holds, does the condition about $\phi$ change?

